Given the family of lines,$a(2x + y + 4) + b(x - 2y - 3) = 0$. Among the lines of the family,the number of lines situated at a distance of $\sqrt{10}$ from the point $M(2, -3)$ is:

For $a > b > c > 0$,the distance between $(1,1)$ and the point of intersection of the lines $ax + by + c = 0$ and $bx + ay + c = 0$ is less than $2\sqrt{2}$. Then:

The equation of the line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes is given by

The point $(4,1)$ undergoes the following transformations successively:
$(i)$ Reflection in the line $x-y=0$
(ii) Shifting through a distance of $2$ units along the positive $X$-axis
(iii) Projection on the $X$-axis
The coordinates of the point in its final position are

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1, -2)$,then the equation of line $BC$ is

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is: